1. Dynamics: The study of the relationships that exist between forces and the motion of objects. Kinematics is the study of motion without regard to forces.

2. Gravitational Nuclear A force of attraction between all masses Electromagnetic Electric forces that exist between stationary charged particles Weak Strong Forces within the particles nucleus Forces that hold the nucleus together.

3. Once in motion, does an object stop moving on its own? Why or why not. Can an object move without a force? Does an apple falling from a tree require a force to move? Will a ball stop moving across a flat frictionless surface?

4. Force: A push, pull or lift. Force is a vector quantity. SI Unit = “N” Newton, direction (1 N = kg·m / s 2 ) Force = (mass)(acceleration); F = ma Field Force  Any force that causes an object to move without the objects being in contact. Ex. The gravitational force, electromagnetic forces. Contact Force  Any force that causes or hinders motion between two objects that are physically touching. Weight (F w )  The measurement of the gravitational force acting on an object. F w = mg = (mass x acceleration due to gravity) F w is perpendicular to the center of t he Earth FwFw

5. Normal Force (F N )  The normal (supporting) force acts perpendicular to the surface of contact and is equal and opposite to the force weight for objects on a horizontal surface.  F N is not equal and opposite to the F w if sitting on an inclined surface. FNFN Force applied (F A )  is any force that causes motion in the same direction as the force applied. Resistive force (F r )  is any force that impedes the motion in the direction of the applied force.

6. Net Force (F net )  is the resulting force after evaluating all the applied and resistive forces acting on the object. F net = F A + F r FYI.. Since resistive forces oppose motion, they are often considered negative, thus the equation turns out to be F net = F A + (- F r ) = F A - F r. If the forces are acting in the same direction you can just add them and let the direction determine the sign. (+. -) When forces are acting a right angles to each other, use the Pythagorean Theorem. When forces are acting at a non-right angles, use the Law of Cosines and the Law of Sines or the “x”, “y” resolution system. The net force is the resultant force which describes the motion.

7. Drawing Free-Body Diagrams (Force Diagrams) Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation The size of the arrow in a free-body diagram is reflective of the magnitude of the force. The direction of the arrow indicates the direction in which the force is acting. Each force arrow in the diagram is labeled to indicate the exact type of force. Generally in a free-body diagram we represent the object by a box and draw the force arrow(s) from the center of the box outward in the direction that the force is acting. You may choose to replace the box with a picture of the object.

8. After looking at the free body diagram (force diagram), in what direction is the object moving? F N = Normal force F A = Force applied F r Resistive force F w = Force of Weight The object is moving to the right since the applied force is greater than the resistive force ∙

9. Given the mass of the object at 50.0 kg, what is the force of weight and the Normal force? F N Normal force F A = Force applied F r Resistive force F w Force of Weight The force of weight is calculated using mass x acceleration due to gravity (g). F w = mg = (50.0 kg)(-9.81 m/s 2 ) F w = -491 N (The negative sign indicates direction.) The Normal force is equal in magnitude but opposite in direction to the force of weight on a horizontal surface. Thus F N = 491 N

10. Given the applied force on this 50.0 kg object is 250N and the resistive force is 160N,what is the net force and how fast is the object accelerating? F N 491 N F A = 250 N F r = -160 N F w -491 N Since all forces = ma, the net force = mass x net acceleration, thus F net = m a net : 90 N = (50.0 kg)(a net ) The net force is calculated using F A + F r = F net, thus 250 N – 160 N = 90 N F net = 90 N a net = 90 N = 1.8 m/s 2 50.0 kg

11. F r = air resistance (10 N) F w = force of weight 27 N F A = wind (8 N) Is this object in freefall? No… Free fall by definition disregards all resistive forces. Find the resultant acceleration. Since motion is occurring in Two directions, determine Your x and y vectors then Use the Pythagorean Theorem and Tangent Functions to determine the resultant. Is there a normal force present? No, because the object is not being supported by another surface.

12. F r = air resistance (10. N) F w = force of weight 27 N The F w is also an applied force in this situation. F A = wind (8.0 N) The vertical force is F r – F w = F y 10. N – 27 N = - 17 N The horizontal force is 8 N The resultant force is the net force in this case, thus [(-17 N) 2 + (8.0N) 2 ] 1/2 = 19 N (-17N) (8.0 N) F net = 19 N (We always add vectors tail to head.)

13. F r = air resistance (10. N) F w = force of weight 27 N F A = wind (8.0 N) The direction in force diagrams is generally relative to the horizontal. We will find the angle where the resultant begins. Tan –1 8.0 N = 25º, 17 N 90º – 25º = 65º below the horizontal (-17N) (8.0 N) F net = 19 N,65º below the horizontal  

14. A 50.0 kg crate is on surface at a 35º inclined plane. Draw a force diagram and calculate the values for all the forces acting on the crate. FwFw FNFN 35 35º The normal force is calculated by looking at the perpendicular force: the force of weight component equal in magnitude but opposite in direction to the normal force. Cos  = F F w -491 Cos 35 = -407 N F w Cos  = F The force of weight is directed toward the Earth’s center. Calculate the force of weight. The normal force is perpendicular to the surface of contact and directed upwards. F w = mg =-491N No, because the object is sitting on a inclined surface. Is the normal force equal to the force of weight? FF

15. A 50.0 kg crate is sitting on a 35º inclined plane. Draw a force diagram and calculate the values for all the forces acting on the crate. FF FwFw FAFA FNFN 35º 35 The Applied force / parallel force is calculated by using the Sine function. Sin  = F || F w -491Sin 35 = -282 N ±F || = ±F A : F A/|| = -282 N F w Sin  = F || The force of weight is acting on the object in another way. It is also trying to pull the box down the ramp. This can be looked at as an applied force. This can also be called the parallel force (F || ) and provides the closing leg of the F w right triangle. F ||

16. FF F w FAFA FNFN F 35º If the box is accelerating down the ramp, the applied force will be greater than the friction force. If the box is moving at constant velocity, the applied and friction force will be equal. What does this force diagram suggest? That’s right, it suggest the box is accelerating down the ramp. It also suggests there is no friction or resistive force present. Is that possible? 35º Thus we must include the resistive force of friction in our diagram. No. Friction is a always present when two object are in contact. F ||

17. Determine the components of a 10. kg mass sitting on the following inclined surfaces. A ramp inclined 15°, 25°, 35°, 45°, 55 Data: A 10. kg mass has a force of weight of 98.1 N 98 Sin 15 = 2598 Cos 15 = 95 98 Sin 25 = 4198 Cos 25 = 89 98 Sin 35 = 5698 Cos 35 = 80. 98 Sin 45 = 6998 Cos 45 = 69 98 Sin 55 = 8098 Cos 55 = 56 Draw some conclusions based on your calculations.

18. Data: A 10. kg mass has a force of weight of 98.1 N (F w = mg) 98 Sin 15 = 25 98 Cos 15 = 95 98 Sin 25 = 41 98 Cos 25 = 89 98 Sin 35 = 56 98 Cos 35 = 80. 98 Sin 45 = 69 98 Cos 45 = 69 98 Sin 55 = 80 98 Cos 55 = 56 Draw some conclusions based on your calculations. 15 166.0 1311 9.0 1. As the ramp’s incline increased, the parallel force also increased. The parallel force’s increase slowed as the angle’s measure grew by regular intervals of 10° 2. As the ramp’s incline increased, the perpendicular (normal) force decreased. The normal force’s decreased more rapidly than the angle’s measure which grew by regular intervals of 10°. 3. At the inclined angle of 45°, the parallel and perpendicular forces were the same. 11 13

19. Inertia: · The ability of an object to maintain it’s present state of motion ….or…. ·The ability of an object to resist a change in motion. Newton's First Law of Motion  The Law of Inertia An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force. …..or….. An object will maintain its present uniform motion until acted upon by another force.

20. Equilibrium occurs when the sum of forces acting at a point is zero. The Equilibrant is equal in magnitude to the resultant, but opposite in direction. Component forces, (f x, f y ) when combined give us the resultant.

21. What happens to you if you hit a deer head-on and you are not wearing a set belt? What happens to a bag of groceries sitting on your car seat when you slam on the breaks? Explain your answer. Newton’s 1 st law of Motion

22. Concepts surrounding Newton’s 1 st Law Forces are balanced thus equilibrium is achieved. Two types of Equilibrium Dynamic equilibrium Objects are moving at constant velocity. Static equilibrium Objects are at rest. To evaluate objects in equilibrium: 1)Sum up all vertical and horizontal forces.  vertical forces:  F y = 0 Vertical forces always include the force of weight.  horizontal forces =  F x = 0 2) Net Force = 0

23. 120 kg 30 ° What tension exists in the top wire ∑F y = F up – F w = 0 ∑F x = F left – F right = 0 F up – F w = 0 F T Sin 30 – mg = 0 F T Sin 30 = mg F T = mg sin 30

24. Given the following picture… Happy Thanksgiving 16° 30° First Draw the Force Diagram ∑F y = F up – F w = 0 f T1

25. Given the following picture… Merry Christmas 16° 30°

26. Newton’s Second Law of Motion Newton's second law of motion pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. The acceleration depends directly depends inversely upon the upon the object’s “Net Force” “mass” Forces are Unbalanced There is an acceleration

27. The acceleration of an object is produced by a net force is directly proportional to the magnitude of the net force, Is in the same direction as the net force, and Is inversely proportional to the mass of the object. the net force is equated to the product of the mass times the net acceleration. F net = m · a net In terms of an equation

28. Sum of Forces Recall: The net force is determined by looking at the sum of all forces  vertical forces =  F y – Vertical forces always include the force of weight.  horizontal forces =  F x Likewise…the net acceleration can be determined by looking at the sum of all the accelerations or by finding the net force and dividing out the mass of the moving object. F net  m = a net F A – F R = F net

29. FF FwFw FAFA FNFN F 20º Cos  = F F w (50.0 kg)(-9.81 m/s 2 ) Cos 20 = -461 N F N = -F  : F N = 461 N F w Cos  = F Equilibrium occurs when the sum of forces acting at a point is zero. The Equilibrant is equal in magnitude to the resultant, but opposite in direction. Component forces, (f x, f y ) when combined give us the resultant.

30. 1. 65N 88N 65N 88N R 2. 1 st Law problems 7.5x10 2 N 9.5x10 2 N R 7.5x10 2 N 9.5x10 2 N What is angle C? 105 ° 3. 600N 750N 675N R A= 600N, N B= 750 N, E C =675N, 120 ° A B C ΣF x = A x + B x + C x ΣF y = A y + B y - C y ΣF x = 0 + 750N + 675Cos30ΣF y = 600N + 0- 675Sin30 ΔXΔX Δ y

31. Atwoods machine Atwood's machine is a device where two masses, M and m, are connected by a string passing over a pulley. Assume that M > m. What is the acceleration of the two masses? Start with a good free-body diagram as shown. Two, in fact, one for each mass. Assume the pulley is frictionless and massless, which means the tension is the same everywhere in the string. We'll learn how to account for the pulley later. T 1 = T 2 = T

32. Mg – (mg + ma) = Ma :Mg – mg – ma = Ma Mg – mg – ma = Ma : Mg – mg = Ma + ma g (M - m) = a (M + m) g = a (M + m) (M – m) For mass m: ∑ F y = ma y T - mg = ma For mass M: ∑F y = Ma y Mg - T = Ma Think about what the system will do. If the system is released from rest, the heavy mass will accelerate down and the lighter one will accelerate up. Align the coordinate systems with the acceleration. Each mass has its own coordinate system, but they must be consistent. Take y down for mass M. Take y up for mass m. Recognize that the masses have the same acceleration, a. Apply Newton's second law for each mass. Combine the equations to eliminate T. T = mg + ma, so:

33. Friction “Friction is a Force that always pushes against an object when it touches another object” “When 2 things are in contact with each other, there will be friction acting between them”

34. Friction… High friction (lots of friction) – will slow something down Low friction (not much friction) – will keep things moving

35. High friction or Low friction? Ski’s on the snow Car tire Brakes on a bike Water on a slide Pencil vs rubber-eraser

36. How can we reduce the friction between 2 objects? Reduce the contact area by using rollers/ball- bearings/wheels Change the surfaces of the materials that are touching by using lubrication eg. Oil Create a cushion of air Eg. Like a hovercraft or air hockey table

37. “Fluid” Friction This type of friction is what happens with liquids and gases (In Physics, liquids and gases are both called "fluids". They behave in similar ways.) Fluid friction is also known as "drag". On aircraft it's also called "air resistance". It depends on:- – how thick the fluid is (its "viscosity") – the shape of the object – the speed of the object

38. Aircraft and car designers want to reduce drag, so that the vehicle can go fast without having to waste too much fuel. How do you think they can do it?

39. Paper “Aeroplane” Competition Aim: The winning plane is the one who gets their plane to travel the farthest, is in the air the longest and thus is the fastest. Materials: 1 sheet of blank paper 1 paperclip Scotch tape Instructions: Use only the materials indicated in the above list; you don’t have to use them all. The plane may be constructed using all or part of the 21.5 x 28 cm sheet of paper. Label each plane with the "engineer’s" name. Each person is allowed three throws; record the best throw only. Begin timing with the throw of the plane and end when the plane hits the floor. Settle a tie with a throw-off. www.worldofteaching.com

40. Friction -the direction of the force is parallel to the surface and opposite the direction of the sliding - the symbol of the friction force is F f - there are two types of friction forces that we will be discussing; Kinetic and static friction force -In almost all situations, static friction is a stronger force than kinetic friction. Kinetic Friction Friction when an object is moving over a surface F k = µ k F N F k = µ k mgCos Ɵ Static Friction Friction when an object is stationary. The friction you must overcome to get an object to move. F s = µ s F N F s = µ s mgCos Ɵ

41. coëfficient of friction - a constant that relates the frictional force on an object to the normal force - The coëfficient of friction is represented by the Greek letter μ (mu). -It is a dimensionless number, which means that it has no units. (This is because μ is a ratio of two forces, so the units cancel.) coëfficient of static friction: μ s represents the coëfficient of friction for an object that is stationary (not moving) coëfficient of kinetic friction: μ k represents the coëfficient of friction for an object that is moving. -The coëfficient of friction takes into account the surface areas and surface characteristics of the objects in contact.

42. Note that the force of static friction is an inequality. For a stationary object, the force that resists sliding is, of course, equal to the force applied. However, once the applied force exceeds μ s F N, the object starts moving and the equation for kinetic friction applies. Friction as a Vector Quantity Like other forces, the force of friction is a vector. Its direction is opposite to the direction of motion. -for an object that is stationary  the force of friction is in the direction opposite to the applied force -for an object that is moving  the force of friction is in the direction opposite to the velocity -This is perpendicular to the normal force, but the direction of the normal force cannot tell us the direction of the force of friction. -This means that whether the force of friction should be positive or negative needs to be determined directly from the coördinate system chosen for the problem.

43. 1.A person pushes a box across the floor at constant velocity. 2.A person pushes a box across the floor at a constant acceleration. 3.A person pushes a box but it does not move. Write down what you know based on these statements. #1: Constant velocity indicates equilibrium  Newton’s 1 st Law  acceleration is Zero  F net = 0 The force applied is equal to the force of friction (resistive forces) kinetic friction is at work #2 Constant acceleration indicates unbalanced forces  Newton’s 2 nd Law  acceleration is > Zero  F net > 0 The force applied is greater than the force of friction (resistive forces) kinetic friction is at work

44. 3.A person pushes a box but it does not move. Static friction is greater than the force applied leading to static equilibrium. F net = 0, object is at rest.

45. Aerodynamic Drag Understand to correctly use the term “drag” your are referring to an object being slowed down by a fluid. Most of the physics problems that involve aerodynamic drag fall into two categories: 1. The drag force is small enough that we ignore it. 2. The drag force is equal to some other force that we can measure or calculate. For simple situations involving aerodynamic drag, the drag force is given by the following equation: F D = -½ ρ v 2 C D A where: F D = drag force ρ = density of the fluid v =velocity of the object C = drag coëfficient of the object (relative to the fluid) (based on its shape) A= cross-sectional area of the object

46. This equation applies when the object has a blunt form factor, and the object’s velocity relative to the properties of the fluid (such as viscosity) causes turbu- lence in the object’s wake (i.e.,behind the object). The drag coëfficient, C D, is a dimensionless number (meaning that it has no units) that encompasses all of the types of friction associated with aero- dynamic drag. It serves the same purpose in drag problems that the coëfficient of friction, μ, serves in problems involving friction between solid surfaces. Approximate drag coëfficients for simple shapes are given below, assuming that the fluid motion relative to the object is in the direction of the arrow.